## Molar Mass of Gas

#### Learning Objective

• Apply knowledge of molar mass to the Ideal Gas Law

#### Key Points

• The molecular weight (molar mass) of any gas is the mass of one particle of that gas multiplied by Avogadro’s number (6.02 x 1023).
• Knowing the molar mass of an element or compound can help us stoichiometrically balance a reaction equation.
• The average molar mass of a mixture of gases is equal to the sum of the mole fractions of each gas (xi) multiplied by the molar mass (Mi) of that particular gas: $\bar { M} =\sum _{ i }^{ }{ { x }_{ i }{ M }_{ i } }$ .

#### Terms

• stoichiometrythe study and calculation of quantitative (measurable) relationships of the reactants and products in chemical reactions (chemical equations)
• molar massthe mass of one mole of an element or compound
• ideal gasa hypothetical gas whose molecules exhibit no interaction and undergo elastic collision with each other and with the walls of the container

## Molar Mass of Gases and Gas Mixtures

Molar mass (M) is equal to the mass of one mole of a particular element or compound; as such, molar masses are expressed in units of grams per mole (g mol–1) and are often referred to as molecular weights. The molar mass of a particular gas is therefore equal to the mass of a single particle of that gas multiplied by Avogadro’s number (6.02 x 1023 ). To find the molar mass of a mixture of gases, you need to take into account the molar mass of each gas in the mixture, as well as their relative proportion.

The average molar mass of a mixture of gases is equal to the sum of the mole fractions of each gas, multiplied by their respective molar masses:

$\bar { M} =\sum _{ i }^{ }{ { x }_{ i }{ M }_{ i } }$

The molar volumes of all gases are the same when measured at the same temperature and pressure (22.4 L at STP), but the molar masses of different gases will almost always vary.

## Calculating Molar Mass using the Ideal Gas Equation

The molar mass of an ideal gas can be determined using yet another derivation of the Ideal Gas Law: $PV=nRT$.

We can write n, number of moles, as follows:

$n=\frac{m}{M}$

where m is the mass of the gas, and M is the molar mass. We can plug this into the Ideal Gas Equation:

$PV=\left(\frac{m}{M}\right)RT$

Rearranging, we get:

$\frac{PV}{RT}=\frac{m}{M}$

Finally, putting the equation in terms of molar mass, we have:

$M=\frac{mRT}{PV}$

This derivation of the Ideal Gas Equation is useful in determining the molar mass of an unknown gas.

## Example

• An unknown gas with a mass of 205 g occupies a volume of 20.0 L at standard temperature and pressure. What is the molar mass of the gas?
• $M=\frac{mRT}{PV}$

$M=\frac{(205)(0.0821)(273)}{(1.0)(20.0)}=230\frac{g}{mol}$